Question

Water flows in a stream line manner through a capillary tube of radius a. The pressure difference being $P$ and the rate of flow is $Q$. If the radius is reduced to $\frac{a}{4}$ and the pressure is increased to $4P$, then the rate of flow becomes

• A. $4Q$
• B. $\frac{Q}{2}$
• C. $Q$
• D. $\frac{Q}{64}$

Answer 1

Answer: (D)

Rate of flow $\propto$ pressure difference $\times$ (radius) ${}^4$
$\left\{\because Q=\frac{\pi P{r}^4}{8\eta L}\right\}$
$Q\propto P\times a^4$
So, $\frac{Q_1}{Q_2}=\frac{P_1{a}_1^4}{P_2{a}_2^4}$
$\frac{Q_1}{Q_2}=\frac{P\times a^4}{4P\times {\left(\frac{a}{4}\right)}^4}=\frac{64}{1}$
$\therefore Q_2=\frac{Q_1}{64}=\frac{Q}{64}$